The greatest integer ≤ Y is denoted by [Y] , and {Y} = Y - [Y].

How many distinct real Y satisfy this equation, whenever 1 ≤ Y ≤ 28 ?

                                         {Y²} = {Y}²

Note: While a solution may be trivial with the aid of a computer program, show how to derive it without one.

Write Y = N + x where N is an integer and 0 <= x < 1. Then [Y] = N and {Y} = x. Now substitute:

{Y^2} = Y^2 - [Y^2] and {Y}^2 = x^2

Y^2 = N^2 + 2Nx + x^2
[Y^2] = [N^2 + 2Nx + x^2] = N^2 + [2Nx] (since integers can be safely removed from inside [] without altering the result.)

Note that this ignores the potential contribution to [] of x^2. Certainly, x^2 < 1, so [2Nx + x^2] either = [2Nx] or [2Nx] + 1. Hold that thought...

For equality to hold, then,

N^2 + 2Nx + x^2 - N^2 - [2Nx] = x^2 or, rearranging terms,

2Nx = [2Nx]

(Remember that thought you were holding? Well, if x^2 matters, then the above equation is instead 2Nx = 1 + [2Nx]. But by definition,
2Nx = [2Nx] + z where 0<=z<1 so for this equation to hold, z would have to be = 1 which is inconsistent with the definition of [].
Therefore, there are no solutions when x^2 "tips the scales", and all solutions come when this term is ignored.)

Only whole numbers have the property that [I] = I so 2Nx = I for some whole number I

For each value of N, then, there are (2N-1) values of x that are solutions: 1/(2N) up to (2N-1)/2N

(Clearly whenever Y is an integer, x=0, and {Y^2} = {Y} = 0 is also a solution but we'll add those in later.)

The values of N that can have nonzero x which are permitted are 1..27 by the initial conditions, so the total number of solutions is

Sum(2i-1) i=1..27 = 27^2 (since the sum of the first N odd numbers = N^2)

Additionally, all of the integral values of Y are solutions, and there are 28 of these, none of which have been thus far counted,
so the overall number of solutions is 27^2 + 28 = 757

These solutions are specifically N + i/2N where 1 <= N < 28 and 1 <= i < N, N and i both integers, as well as the integers from 1 to 28 inclusive.

As a test, consider N=3, i=2 giving the solution 3+2/6 = 10/3

{10/3} = 1/3. (10/3)^2 = 100/9 = 11 + 1/9 so {(10/3)^2} = 1/9

(1/3)^2 = 1/9 so this is indeed a solution. Naturally, these solutions are all unique. Since if A = B for some A and B then [A] = [B],
and [N + i/2N] = N, two equal solutions requires that both have the same value of N or else their [] will not match. But if they have
the same N, then they also have the same i and hence are identical after all, so there is no double-counting in the overall solution count.

Posted by Paul on 2008-09-06 22:31:54